Suppose that a queue has $N$-policy, that is, only when there are $N$ customers in the queue the service starts. In such a queue with deterministic arrival rate $\lambda$ and deterministic service rate $\mu$ (Nothing is stochastic here and inter arrival time is more than service time),
(a) what will be the waiting time of all the customers (from the time of start of service) until the queue size is zero for the first time?
(b) In how much time, will the queue be of size zero for the first time?

The waiting time of the $N$ customers is easy to figure out: $$\frac{N(N+1)}{2\mu}.$$ The rest am not sure about. Kindly help me with the two questions.

  • $\begingroup$ Formatting tips here. $\endgroup$
    – Em.
    Commented Feb 11, 2016 at 7:52
  • $\begingroup$ I am unsure what is exactly asked in (a) and (b). In (a), is it the time to go from $N$ customers to $0$ customers? In (b), is it the time to go from $0$ to $0$ customers? $\endgroup$
    – Ritz
    Commented Feb 12, 2016 at 12:22
  • $\begingroup$ (b) is the time for going from $N$ customers to $0$ customers. (a) is the sum of waiting time of each of the customers from the time the service starts (when there are $N$ customers waiting) till the time the service ends (the first time there are $0$ customers in the system). $\endgroup$
    – Insoucyant
    Commented Feb 15, 2016 at 3:03

1 Answer 1


The upward arrows denote arrival while he downward arrows denote departure. Both arrivals and departures happen at every $\kappa \tau $ time units from the start of the service (as well at every time unit which is L.C.M. of $\kappa$ and $\tau$). In between all multiples of $\kappa \tau $ the queue length shrinks by $\kappa - \tau $ units, till it becomes $0$. The time between the start of the service till the queue is of size $0$ for the first time since the start of the service will be: $$\frac{\kappa \tau (N-1)}{\kappa - \tau} + \tau = \frac{(N-1)}{\mu - \lambda} + \frac{1}{\mu}$$

To calculate the total backlogging cost during the service period, let: \begin{align*} \kappa &= a \alpha \\ \tau &= a \beta \end{align*} where $a$ is the G.C.D. of $\kappa$ and $\tau$. Let $t_1, t_2, \dots , t_n $ be multiples of $a$ where $ t_n = \frac{\kappa \tau (N-1)}{\kappa - \tau} $ and $n = N-1$. Between every $t_i$ and $t_{i-1}, (1<i<N+1)$ each arrival will find the same queue length, hence the waiting time for all such arrivals will be equal. The number of arrivals between each $t_i$ and $t_{i-1}$ will be $(\alpha - \beta)$.\ The demands that are generated at $t_1, t_2, \dots , t_n $ will find the queue of length $(N-1), \dots, 1$ respectively at the time of arrival. The total backlogging cost of all the demands that are generated during the service period will be :- \begin{equation*} \textit{Backlogging $Cost_S$/Inner Cycle} = \frac{N(N-1)}{2 \mu} (\alpha - \beta +1) \pi \end{equation*}



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